Courses
Courses for Kids
Free study material
Offline Centres
More Last updated date: 26th Nov 2023
Total views: 279.6k
Views today: 4.79k

# The sum of the first n terms of the series ${{1}^{2}}+{{2.2}^{2}}+{{3}^{2}}+{{2.4}^{2}}+{{5}^{2}}+{{2.6}^{2}}+\cdots$ is $\dfrac{n{{\left( n+1 \right)}^{2}}}{2}$, when n is even. When n is odd sum is: Verified
In the question, we are given series as ${{1}^{2}}+{{2.2}^{2}}+{{3}^{2}}+{{2.4}^{2}}+{{5}^{2}}+{{2.6}^{2}}+\cdots$and we also know that the sum of the series is $\dfrac{n{{\left( n+1 \right)}^{2}}}{2}$where n is even this is provided.
Now, we need to find the sum of the same series but when n is odd. So, for that we need to let $n=2m+1$ . Now, the required sum would be \begin{align} & {{1}^{2}}+{{2.2}^{2}}+{{3}^{2}}+{{2.4}^{2}}+{{5}^{2}}+{{2.6}^{2}}+\cdots +2{{\left( 2m \right)}^{2}}+{{\left( 2m+1 \right)}^{2}} \\ & \Rightarrow \sum{{{\left( 2m+1 \right)}^{2}}+4\left( {{1}^{2}}+{{2}^{2}}+\cdots +{{m}^{2}} \right)} \\ \end{align}
\begin{align} & =\dfrac{\left( 2m+1 \right)\left( 2m+2 \right)\left( 4m+2+1 \right)}{6}+\dfrac{4m\left( m+1 \right)\left( 2m+1 \right)}{6} \\ & \Rightarrow \dfrac{\left( 2m+1 \right)\left( m+1 \right)}{6}\left[ 2\left( 4m+3 \right)+4m \right] \\ & \Rightarrow \dfrac{\left( 2m+1 \right)\left( 2m+2 \right)\left( 6m+3 \right)}{6} \\ & \Rightarrow \dfrac{{{\left( 2m+1 \right)}^{2}}\left( 2m+2 \right)}{2} \\ \end{align}
Now, we have almost attained the answer, further simplifying this would give us the final answer which is $\dfrac{{{n}^{2}}\left( n+1 \right)}{2}$ .
Therefore, the sum of given series when the number of terms is odd is given by $\dfrac{{{n}^{2}}\left( n+1 \right)}{2}$.